a) To show that there exists a polynomial $Q(x)$ such that $P(x) = (x-a)^2Q(x)$, we can use the Factor Theorem. Since $P(a) = P'(a) = 0$, it means that $(x-a)$ is a factor of both $P(x)$ and $P'(x)$. Therefore, we can write $P(x)$ as $P(x) = (x-a)Q(x)$ for some polynomial $Q(x)$. Now, differentiating both sides of this equation, we get $P'(x) = Q(x) + (x-a)Q'(x)$. Since $P'(a) = 0$, we have $Q(a) = 0$. So, $(x-a)$ is also a factor of $Q(x)$. Therefore, we can write $Q(x)$ as $Q(x) = (x-a)R(x)$ for some polynomial $R(x)$. Substituting this into our previous equation for $P(x)$, we get $P(x) = (x-a)^2R(x)$. Thus, there exists a polynomial $Q(x)$ such that $P(x) = (x-a)^2Q(x)$.
b) To show that there exists at most one line $\ell$ that is tangent to the graph of $P(x)$ at two places, we can use the fact that a quartic polynomial can have at most four distinct roots. If a line $\ell$ is tangent to the graph of $P(x)$ at two places, it means that the line intersects the graph at those two points but does not cross it. Since a quartic polynomial can have at most four distinct roots, and each root corresponds to an intersection point with the $x$-axis, there can be at most two tangent points. Therefore, there exists at most one line $\ell$ that is tangent to the graph of $P(x)$ at two places.